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The real sequence $x_n$ satisfies $x_{n+1}=2x_n^2-1$ and $x_1=\frac45$

I want to know if this sequence convergences.

I think it converges to $-\frac12$; considering the graphs of $y=x$ and $y=2x^2-1$

Is it right? If so, how to prove that?

I tried to find a positive number $c$ $(0\lt c \lt1)$ such that $\forall n,\left | x_{n+2}-x_{n+1} \right | \leq c \left| x_{n+1}-x_{n} \right |$ which would prove this sequence is cauchy, but I failed.

gebruiker
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Hwan
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  • Related : http://math.stackexchange.com/questions/781024/for-x-n1-x-n2-2-show-lim-n-to-infty-fracx-nx-0x-1-cdots-x-n-1-2/781156#781156 –  May 16 '14 at 13:47
  • I have no idea... I know $x_n=cos{\theta 2^{n-1}} , cos\theta=\frac{4}{5}$(in fact i got sequence form this.) But how to prove convergence of it? – Hwan May 16 '14 at 14:06

2 Answers2

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Let $f(x)=2x^2-1$. If the sequence converges then the limit is a fixpoint of $f$, i.e. either $+1$ or $-\frac12$. By induction, we have $|x_n|\le 1$ and $x_n\in\mathbb Q$ for all $n$. One way of convergence would be that the sequence already lands on a fixed point after finitely many steps. But it could land on $+1$ only via $$ \tag1\ldots\to \pm\frac{\sqrt2}2\to 0\to -1\to1\to1\to\ldots$$ and it could land on $-\frac12$ only via $$\tag2 \ldots\to\pm\frac{\sqrt{3}}2\to \frac12\to-\frac12\to-\frac12\to\ldots,$$ which is impossible as all $x_n$ are rational (and $x_1\notin\{-1,-\frac12,0,\frac12,1\}$). Note that for $x\ne 1$ we have $$\tag3 \frac{f(x)-1}{x-1}=2(x+1)>1 \quad\text{if }x>-\frac12$$ and for $x\ne-\frac12$ $$\tag4 \frac{f(x)+\frac12}{x+\frac12}=2x-1<-1 \quad\text{if }x<0.$$ We conclude that the sequence moves away from the fixpoints as soon as it is somewhat close to them. Hence no convergence.

Remarks:

  1. We would have convergence with $x_1\in\{-1,-\frac12,0,\frac12,1\}$ and with any of the countably many irrational preimages that would extend $(1)$, $(2)$ to the left. As other hints suggest, these are the values $x_1=\cos(\alpha)$ such that $\cos(2^n\alpha)=1$ or $\cos(2^n\alpha)=-\frac12$ for some $n\in\mathbb N_0$.
  2. The repelling property $(3)$ and $(4)$ is a consequence of $|f'(x)|=|4x|>1$ for the fixpoints $x=1$ and $x= -\frac12$.
Did
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Hagen von Eitzen
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  • Nice answer! very helpful! – Hwan May 16 '14 at 14:51
  • Sorry i can't understand why it moves away form fixpoints... what's the meaning of the below inequality? and is it contradiction that it would have convergence if it moves away the fixpoints? – Hwan May 16 '14 at 15:06
  • For $x_n$ close enough (as we see, $|x_n-c|<\frac12$ is sufficient) to the assumed limit $c$, we have that $\frac{|x_{n+1}-c|}{|x_n-c|}>1$, i.e. $|x_{n+1}-c|>|x_{n}-c|$. If $x_n\to c$ then there exists $N=N(\frac12)$ such that for all $n\ge N$, we have $|x_n-c|<\frac12$. But then $|x_n-c|>|x_N-c|$ for all $n>N$. Hence for $\epsilon:=|x_N-c|>0$(!) we cannot find a suitable $N(\epsilon)$. – Hagen von Eitzen May 16 '14 at 19:47
  • Change variables with $x_n = \dfrac{y_n}{2}$ and you get $y_{n+1}=y_n^2-2$, whose Julia set is the interval $[-2,2]$. Therefore, even though the sequence does not converge, it remains bounded forever and its image is dense on $[-1,1]$. Just a fun fact. – lhf Mar 29 '16 at 11:23
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Hint,Note $$\cos{(2x)}=2\cos^2{x}-1$$

math110
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